算法学步 PAT A1012.The Best Rank

题目

To evaluate the performance of our first year CS majored students, we consider their grades of three courses only: C - C Programming Language, M - Mathematics (Calculus or Linear Algrbra), and E - English. At the mean time, we encourage students by emphasizing on their best ranks -- that is, among the four ranks with respect to the three courses and the average grade, we print the best rank for each student.
For example, The grades of C, M, E and A - Average of 4 students are given as the following:

StudentID  C  M  E  A
310101     98 85 88 90
310102     70 95 88 84
310103     82 87 94 88
310104     91 91 91 91

Then the best ranks for all the students are No.1 since the 1st one has done the best in C Programming Language, while the 2nd one in Mathematics, the 3rd one in English, and the last one in average.
输入格式
Each input file contains one test case. Each case starts with a line containing 2 numbers N and M (≤2000), which are the total number of students, and the number of students who would check their ranks, respectively. Then N lines follow, each contains a student ID which is a string of 6 digits, followed by the three integer grades (in the range of [0, 100]) of that student in the order of C, M and E. Then there are M lines, each containing a student ID.
输出格式
For each of the M students, print in one line the best rank for him/her, and the symbol of the corresponding rank, separated by a space.
The priorities of the ranking methods are ordered as A > C > M > E. Hence if there are two or more ways for a student to obtain the same best rank, output the one with the highest priority.
If a student is not on the grading list, simply output N/A。

输入样例：

5 6
310101 98 85 88
310102 70 95 88
310103 82 87 94
310104 91 91 91
310105 85 90 90
310101
310102
310103
310104
310105
999999

输出样例：

1 C
1 M
1 E
1 A
3 A
N/A

题意：

现已知n个考生的三门课的分数C、M、E，而平均分A可以由这三个分数得到。现在分别按这4个分数对n个考生从高到低排序，这样对每个考生来说，就会有4个排名且每个分数都会有一个排名。接下来会有m个查询，每个查询输入一个考生的ID，输出该考生4个排名中最高的那个排名及对应是A、C、M、E中的哪一个。如果对不同的课程有相同的排名，则按照优先级A、C、M、E输出；如果查询的考生的ID不存在，则输出N/A。

程序代码（来源：算法笔记）

#include <iostream>
#include <cmath>
#include <cstdio>
#include <algorithm>
using namespace std;
 
struct Student {
    int id; //存放6位整数的ID
    int grade[4];   //存放4个分数
}stu[2010];
 
char course[4] = {'A', 'C', 'M', 'E'}; //按优先级顺序，方便输出
int Rank[10000000][4] = {0}; //Rank[id][0] ~ Rank[id][4]为4门课对应的排名
int now; //cmp函数中使用，表示当前按now号分数排序stu数组
 
bool cmp(Student a, Student b) {    //stu数组按now分数递减排序
    return a.grade[now] > b.grade[now];
}
int main() {
    int n, m;
    scanf("%d%d", &n, &m);
    //读入分数，其中grade[0] ~ grade[3]分别代表A, C, M, E
    for(int i = 0; i < n; i++) {
        scanf("%d%d%d%d", &stu[i].id, &stu[i].grade[1], &stu[i].grade[2], &stu[i].grade[3]);
        stu[i].grade[0] = round((stu[i].grade[1] + stu[i].grade[2] + stu[i].grade[3]) / 3.0) + 0.5;
    }
    for(now = 0; now < 4; now++) {  //枚举A, C, M, E 4个中的一个
        sort(stu, stu + n, cmp);    //对所有考生按该分数从大到小排序
        Rank[stu[0].id][now] = 1;   //排序完，将分数最高的设为rank1
        for(int i = 1; i < n; i++) {    //对于剩下的考生
            //若与前一位考生分数相同
            if(stu[i].grade[now] == stu[i - 1].grade[now]) {
                Rank[stu[i].id][now] = Rank[stu[i - 1].id][now];    //则他们排名相同
            } else {
                Rank[stu[i].id][now] = i + 1;   //否则，为其设置正确的排名
            }
        }
    }
    int query; //查询的考生ID
    for(int i = 0; i < m; i++) {
        scanf("%d", &query);
        if(Rank[query][0] == 0) {   //如果这个考生ID不存在，则输出“N/A”
            printf("N/A
");
        } else {
            int k = 0;  //选出Rank[query][0~3]中最小的（rank值越小，排名越高）
            for(int j = 0; j < 4; j++) {
                if(Rank[query][j] < Rank[query][k]) {
                    k = j;
                }
            }
            printf("%d %c
", Rank[query][k], course[k]);
        }
    }
    return 0;
}